Some questions on minimal log discrepancies

Transcription

1 Some questions on minimal log discrepancies Mircea Mustaţă University of Michigan Aussois June 25, 2015 Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

2 Setting Let X be our ambient variety (assumed smooth or with mild singularities). We work over k = k, with char(k) = 0. Assume dim(x ) 2. Let Z X be defined by some I Z O X. We want to study the singularities of the pair (X, qz), where q Q 0 (or q R 0 ). Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

3 Setting Let X be our ambient variety (assumed smooth or with mild singularities). We work over k = k, with char(k) = 0. Assume dim(x ) 2. Let Z X be defined by some I Z O X. We want to study the singularities of the pair (X, qz), where q Q 0 (or q R 0 ). The invariant that we will discuss is obtained by considering all divisorial valuations of k(x ). Such a valuation corresponds to a prime divisor E Y, where Y normal, with a birational morphism Y X. Get valuation ord E on k(x ) = k(y ). Its center is c X (E) := f (E). Example. If x X smooth point and E is the exceptional divisor on Bl x (X ), then ord E = ord x, where ord x (f ) = max{j f I j x}. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

4 The log discrepancy Given a pair (X, qz) as above and ord E, one measures the singularities of the pair with respect to this valuation by considering q ord E (Z) := q min{ord E (h) h I Z }. These numbers have to be normalized. This is done using the log discrepancy function. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

5 The log discrepancy Given a pair (X, qz) as above and ord E, one measures the singularities of the pair with respect to this valuation by considering q ord E (Z) := q min{ord E (h) h I Z }. These numbers have to be normalized. This is done using the log discrepancy function. Suppose that X is smooth and E is a prime divisor on Y, with f : Y X birational. May assume Y is smooth (replace Y by Y sm ). We have a morphism of vector bundles f (Ω X ) Ω Y of the same rank, which drops rank along a divisor (defined by the determinant of the map), denoted by K Y /X. The log discrepancy of ord E (with respect to X ) is A X (ord E ) := ord E (K Y /X ) + 1. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

6 The log discrepancy, cont d The log discrepancy measures how far the divisor lies over X. Example. If W X smooth subvariety and E is the exceptional divisor on Bl W (X ), then A X (ord E ) = codim X (W ). Remark. If Y Y X are birational morphisms of smooth varieties and E is a prime divisor on Y, then A X (ord E ) = A Y (ord E ) + ord E (K Y /X ). Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

7 The log discrepancy, cont d The log discrepancy measures how far the divisor lies over X. Example. If W X smooth subvariety and E is the exceptional divisor on Bl W (X ), then A X (ord E ) = codim X (W ). Remark. If Y Y X are birational morphisms of smooth varieties and E is a prime divisor on Y, then A X (ord E ) = A Y (ord E ) + ord E (K Y /X ). Since X is smooth, every divisor E can be obtained by a sequence of smooth blow-ups. By the example and remark, A X (ord E ) is determined by the codimensions of the centers and by which proper transforms of previous exceptional divisors contain the center at each step. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

8 Remarks on the singular case When X is singular, there are several possible definitions. In birational geometry the most useful is the following: Assume that X is normal, hence K X makes sense as a Weil divisor. Assume also that X is Q-Gorenstein, that is, some mk X is Cartier, m 1. One can still define K Y /X when f : Y X is birational, with Y smooth, as the unique divisor supported on Exc(f ), linearly equivalent to K Y f (K X ) (note: it might not be effective). Then define A X (ord E ) as before. However: meaning is somewhat more subtle. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

9 Minimal log discrepancy Fix a pair (X, qz) and a closed subset W X (most of the time will take W to be a point). The minimal log discrepancy of (X, qz) with respect to W is mld W (X, qz) := inf {A X (ord E ) q ord E (Z)}. c X (E) W Note: good singularities of (X, qz) small ord E (Z) large mld. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

10 Minimal log discrepancy Fix a pair (X, qz) and a closed subset W X (most of the time will take W to be a point). The minimal log discrepancy of (X, qz) with respect to W is mld W (X, qz) := inf {A X (ord E ) q ord E (Z)}. c X (E) W Note: good singularities of (X, qz) small ord E (Z) large mld. Basic facts: If mld W (X, qz) < 0, then mld W (X, qz) =. Otherwise, one says that (X, qz) is log canonical (in a neighborhood of W ). If (X, qz) log canonical and f : Y X is a monomialization of I Z such that f 1 (W ) is a divisor, then the infimum is a minimum over the divisors on Y. Say: a divisor E computes mld W (X, qz) if c X (E) W and mld W (X, qz) = A X (ord E ) q ord E (Z). Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

11 Comparison with the log canonical threshold It is instructive to compare mld W (X, qz) to another invariant of the pair (X, Z), the log canonical threshold lct(x, Z). This is defined as lct(x, Z) := max{t 0 (X, tz) is log canonical} = min E A X (ord E ) ord E (Z). In spite of the similarity in the definition, the minimal log discrepancy turns out to be a much more subtle invariant than the log canonical threshold. In particular, the analogues of the questions we will see later are now well-understood for log canonical thresholds. This difference is analogous to that between linear programming and integer programming. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

12 Example 1: the monomial case Let Z X = A n defined by monomial ideal I. Let P I be the Newton polyhedron P I = conv{u Z n 0 x u I }. If E u is the toric divisor corresponding to u Z n 0 A X (ord Eu ) = u u n. (primitive), then Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

13 Example 1: the monomial case Let Z X = A n defined by monomial ideal I. Let P I be the Newton polyhedron P I = conv{u Z n 0 x u I }. If E u is the toric divisor corresponding to u Z n 0 A X (ord Eu ) = u u n. (primitive), then It follows that (A n, qz) is log canonical if and only if (1,..., 1) q P I. For such q, we have mld 0 (A n, qz) = min{u u n q min v P I u, v } u Z n >0}. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

14 Example 2: a cone with isolated singularities Let Z = V (f ) A n, with f homogeneous, such that Z has an isolated singularity at 0. Bl 0 A n A n is a monomialization of f, hence If this is the case, then (A n, qz) is log canonical iff q min{1, n/d}. mld 0 (A n, qz) = n qd. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

15 Two open questions Semicontinuity conjecture (Ambro). Given (X, qz), the function X x mld x (X, qz) is lower semicontinuous, that is, all sets {x X mld x (X, qz) t} are open. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

16 Two open questions Semicontinuity conjecture (Ambro). Given (X, qz), the function X x mld x (X, qz) is lower semicontinuous, that is, all sets {x X mld x (X, qz) t} are open. ACC conjecture (Shokurov). Fix a DCC subset Γ of R 0. If n is fixed, then the set satisfies ACC. {mld x (X, qz) x X, dim(x ) n, q Γ} Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

17 Two open questions Semicontinuity conjecture (Ambro). Given (X, qz), the function X x mld x (X, qz) is lower semicontinuous, that is, all sets {x X mld x (X, qz) t} are open. ACC conjecture (Shokurov). Fix a DCC subset Γ of R 0. If n is fixed, then the set satisfies ACC. {mld x (X, qz) x X, dim(x ) n, q Γ} The interest in these comes from the following Theorem (Shokurov). If the above conjectures hold, then we have Termination of Flips in the Minimal Model Program. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

18 A valuation-theoretic version of semicontinuity Using Ambro s work, one can reduce the Semicontinuity Conjecture to the following purely valuation-theoretic conjecture. Here a divisorial valuation is a valuation of the form q ord E, with q Z >0. Semicontinuity conjecture, strong version. Given X affine and two closed irreducible subsets T 1 T 2 X, if v 2 is a divisorial valuation of k(x ) with c X (v 2 ) = T 2, then there is a divisorial valuation v 1 of k(x ) such that: i) c X (v 1 ) = T 1. ii) v 1 (f ) v 2 (f ) for every f O(X ). iii) A X (v 1 ) A X (v 2 ) + codim T2 (T 1 ). Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

19 A valuation-theoretic version of semicontinuity Using Ambro s work, one can reduce the Semicontinuity Conjecture to the following purely valuation-theoretic conjecture. Here a divisorial valuation is a valuation of the form q ord E, with q Z >0. Semicontinuity conjecture, strong version. Given X affine and two closed irreducible subsets T 1 T 2 X, if v 2 is a divisorial valuation of k(x ) with c X (v 2 ) = T 2, then there is a divisorial valuation v 1 of k(x ) such that: i) c X (v 1 ) = T 1. ii) v 1 (f ) v 2 (f ) for every f O(X ). iii) A X (v 1 ) A X (v 2 ) + codim T2 (T 1 ). This is known when X is toric (Ambro) X is smooth (Ein, M-, Yasuda) dim(x ) = 2 (in general, can reduce to X having terminal singularities) Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

20 Outline of proof in the smooth case The proof uses the description of divisorial valuations on smooth varieties via arcs. For the sake of concreteness, say X = A n. Consider the space of arcs of X : X = Hom(Spec k[[t]], X ) = k[[t]] n. A cylinder C in X is the inverse image of a locally closed subset S of (k[[t]]/(t m+1 )) n via the obvious projection map. We put codim X C := codim Xm S. For every C X and f O(X ), we put ord C (f ) := min γ C ord tγ (f ) Z 0 { }. The following result gives an approach to divisorial valuations via cylinders in X. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

21 Outline of proof in the smooth case, cont d Theorem (Ein, M-, Lazarsfeld). Let X be a smooth variety. 1) If C is an irreducible, closed cylinder in X, then ord C extends to a divisorial valuation of k(x ), whose center is the closure of the image of C in X. 2) Given any divisorial valuation v of k(x ), there is a unique maximal irreducible closed cylinder C(v) such that v = ord C(v). 3) We have codim X C(v) = A X (v). What is C(v): say v = q ord E, where E smooth divisor on smooth Y, with f : Y X birational. We have C(v) = f (Cont m (E)). Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

22 Outline of proof in the smooth case, cont d Goal: explain how the previous result implies the valuation-theoretic version of semicontinuity when X is smooth. Recall that v 2 is a divisorial valuation with center T 2 and T 1 is an irreducible closed subset of T 2. Let π : X X be the canonical projection and consider an irreducible component C C(v 2 ) π 1 (T 1 ) that dominates T 1. Then v 1 = ord C satisfies our requirements: C dominates T 1 implies c X (v 1 ) = T 1. C C(v 2 ) implies v 1 = ord C ord C(v2 ) = v 2. A X (v 1 ) = codim X (C(v 1 )) codim X (C) codim X (C(v 2 )) + codim T2 (T 1 ) = A X (v 2 ) + codim T2 (T 1 ). Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

23 A boundedness question on divisors computing mld s We now turn to the ACC conjecture. We will be interested in a special case, when the ambient variety is fixed (this is sufficient, for example, if we are only interested in the case of ambient smooth varieties). What follows is joint work with Yusuke Nakamura. The following question is motivated by the above problem: Question (Nakamura). Let X and x X be fixed, and let also q R >0 be fixed. Is there a fixed M > 0 such that for every subscheme Z X with (X, qz) log canonical, there is a divisor E over X such that a) E computes mld x (X, qz), and b) A X (ord E ) M? Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

24 A boundedness question on divisors computing mld s We now turn to the ACC conjecture. We will be interested in a special case, when the ambient variety is fixed (this is sufficient, for example, if we are only interested in the case of ambient smooth varieties). What follows is joint work with Yusuke Nakamura. The following question is motivated by the above problem: Question (Nakamura). Let X and x X be fixed, and let also q R >0 be fixed. Is there a fixed M > 0 such that for every subscheme Z X with (X, qz) log canonical, there is a divisor E over X such that a) E computes mld x (X, qz), and b) A X (ord E ) M? Remark. It is easy to see that if this has a positive answer, then for every DCC subset Γ R 0, the set satisfies ACC. {mld x (X, qz) q Γ, Z X } Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

25 A partial result Theorem (M-, Nakamura). If X, x X, and q R >0 are fixed, then there a fixed M > 0 such that for every subscheme Z X with (X, qz) log canonical, there is a divisor E over X such that a) E computes mld x (X, qz). b) ord E (I x ) M. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

26 A partial result Theorem (M-, Nakamura). If X, x X, and q R >0 are fixed, then there a fixed M > 0 such that for every subscheme Z X with (X, qz) log canonical, there is a divisor E over X such that a) E computes mld x (X, qz). b) ord E (I x ) M. Remark 1: Suppose that X is smooth and we describe E via a sequence of smooth blow-ups. To give a positive answer to the question: need to bound the length of this sequence. The theorem says that all but a bounded number of the blow-ups in the sequence are free blow-ups (the center is contained in only one proper transform of the previous exc. divisors). Remark 2: A X (ord Em ) ord Em (I x ) lct(x, {x}), hence the assertion in the theorem would follow from a positive answer to the question. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

27 Sketch of proof The proof uses the generic limit construction (de Fernex, M-; Kollár). Say we have a sequence of subschemes Z m X such that each (X, qz m ) is log canonical and no matter how we choose divisors E m computing mld x (X, qz m ), we have lim m ord Em (I x ) =. For simplicity: say X = A n k, x = 0, and Z m is defined by f m k[x 1,..., x n ]. By putting together the coefficients of the f m, we get a formal power series f K[[x 1,..., x n ]], where K/k is a suitable field extension (obtained by a non-standard extension). This has the property that after passing to a subsequence, if X = Spec K[[x 1,..., x n ]] and Z is defined by f, then we may assume lct 0 (X, qz ) = lim m lct 0(X, qz m ). By assumption q lct 0 (X, Z m ) for every m 1, hence q lct 0 (X, Z ). There are two cases to consider. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

28 Sketch of proof, cont d Case 1: mld 0 (X, qz ) = 0. If E is a divisor over 0 X computing this mld, after passing to a subsequence, this comes from a sequence of divisors E m over 0 X such that A X (ord Em ) = A X (ord E ) ord Em (I 0 ) = ord E (I 0 ) and ord Em (f m ) = ord E (f ). This is non-trivial: it uses the adic semicontinuity property of log canonical thresholds. One sees that in this case each E m computes mld 0 (X, qz m ) = 0, while {ord Em (f m )} is bounded, a contradiction. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

29 Sketch of proof, cont d Case 2: mld 0 (X, qz ) > 0. In this case there is δ > 0 such that lct 0 (X, qz + δ{0}) = 1. Key point: after passing to a subsequence, we may assume that Indeed, we may assume that lct 0 (X, qz m + δ{0}) 1 for all m. lim lct 0(X, qz m + δ{0}) = 1 m and the assertion follows from ACC for log canonical thresholds. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

30 Sketch of proof, cont d Let s choose now for each m a divisor E m that computes mld 0 (X, qz m ). By the key point, we have By choice of E m, we have Combining these, we get A X (ord Em ) q ord Em (Z m ) + δ ord Em (I 0 ). A X (ord Em ) q ord Em (Z m ) = mld 0 (X, qz m ). ord Em (I 0 ) mld 0(X, qz m ) δ mld 0(X ). δ The right-hand side is independent of m and this completes the proof of the theorem. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

31 The boundedness question: the monomial case Nakamura s question has a positive answer for monomial subschemes of X = A n : Suppose that Z m = V (I m ), with m 1, are such that each I m k[x 1,..., x n ] is a monomial ideal with (A n, qz m ) log canonical such that no matter how we choose E m that computes mld 0 (A n, qz m ), we have lim m A X (ord Em ) =. For simplicity, assume q Q. One can easily reduce to the case when all I m are (x 1,..., x n )-primary Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

32 The boundedness question: the monomial case Nakamura s question has a positive answer for monomial subschemes of X = A n : Suppose that Z m = V (I m ), with m 1, are such that each I m k[x 1,..., x n ] is a monomial ideal with (A n, qz m ) log canonical such that no matter how we choose E m that computes mld 0 (A n, qz m ), we have lim m A X (ord Em ) =. For simplicity, assume q Q. One can easily reduce to the case when all I m are (x 1,..., x n )-primary By a result of Maclagan, after restricting to a subset and reordering, we may assume that I 1 I 2..., hence mld 0 (A n, qz 1 ) mld 0 (A n, qz 2 )... All these mld s lie in 1 r Z 0 for some integer r 1, hence they stabilize for m m 0. We now see that if E computes mld 0 (A n, qz m0 ), it also computes mld 0 (A n, qz m ) for m m 0, a contradiction. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

33 The boundedness question: the surface case Nakamura s question has a positive answer when X is a smooth surface: Suppose that E is a divisor computing mld x (X, qz) with A X (ord E ) minimal. Say Z = Z(f ). Consider the corresponding sequence of point blow-ups: π N π XN 1... X 2 π 2 1 X1 X0 = X, X N with X i+1 = Bl xi (X i ), with exceptional divisor E i, where x i = c Xi (E). We have E = E N and need to bound N in terms of q. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

34 The boundedness question: the surface case Nakamura s question has a positive answer when X is a smooth surface: Suppose that E is a divisor computing mld x (X, qz) with A X (ord E ) minimal. Say Z = Z(f ). Consider the corresponding sequence of point blow-ups: π N π XN 1... X 2 π 2 1 X1 X0 = X, X N with X i+1 = Bl xi (X i ), with exceptional divisor E i, where x i = c Xi (E). We have E = E N and need to bound N in terms of q. For every i with 1 i N 1, we put p i = q ord xi ( Z), where Z stands for the strict transform of Z. Note that we have 2 p 0 p 1... p N 1 > 0, where the first inequality comes from the fact that (X, qz) is log canonical. The p i lie in a discrete set only depending on q. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

35 The boundedness question: the surface case,cont d Let τ i := A X (ord Ei ) q ord Ei (Z). For each blow-up X i+1 X i we have two cases: Case 1: x i only lies on E i ( free blow-up ). In this case τ i+1 = τ i + 1 p i. Case 2. x i lies on E i and on the strict transform of E j, with j < i. In this case τ i+1 = τ i + τ j p i. By the theorem, the number of blow-ups in Case 2 is bounded. Therefore we only need to show that also the number of blow-ups in Case 1 is bounded. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

36 The boundedness question: the surface case,cont d Let τ i := A X (ord Ei ) q ord Ei (Z). For each blow-up X i+1 X i we have two cases: Case 1: x i only lies on E i ( free blow-up ). In this case τ i+1 = τ i + 1 p i. Case 2. x i lies on E i and on the strict transform of E j, with j < i. In this case τ i+1 = τ i + τ j p i. By the theorem, the number of blow-ups in Case 2 is bounded. Therefore we only need to show that also the number of blow-ups in Case 1 is bounded. Important point: if p i < 1 and we are in Case 1, then τ i+1 > τ i. In fact, there is ɛ > 0 (only depending on q), such that if this is the case, then τ i+1 > τ i + ɛ. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

37 The boundedness question: the surface case,cont d (1) One can show that if p i = p i+1 = 1, then mld x (X, qz) is computed by E i+1, hence i = N 1. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

38 The boundedness question: the surface case,cont d (1) One can show that if p i = p i+1 = 1, then mld x (X, qz) is computed by E i+1, hence i = N 1. (2) Using the fact that (X, qz) is log canonical, we can bound the number of p i > 1, hence the number of those 1. Note that if X i+1 X i is in Case 1, then we still have τ i+1 τ i 1. Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24

39 The boundedness question: the surface case,cont d (1) One can show that if p i = p i+1 = 1, then mld x (X, qz) is computed by E i+1, hence i = N 1. (2) Using the fact that (X, qz) is log canonical, we can bound the number of p i > 1, hence the number of those 1. Note that if X i+1 X i is in Case 1, then we still have τ i+1 τ i 1. (3) Note also that in Case 2, we always have τ i+1 τ i p i 2. Since we have only finitely many steps in Case 2 or with p i 1 and otherwise τ i+1 > τ i + ɛ, and since we conclude that N is bounded. τ N = mld x (X, qz) mld x (X ) = 2, Mircea Mustaţă (University of Michigan) Some questions on minimal log discrepancies Aussois June 25, / 24